Question

PS and PT are tangents drawn to the circle with centre O.
Angle SOT is 120° .
Prove that 4PS square = 3OP square.
Please solve this without trignometry ..With Pythagoras theoremo or any other

Answer 1

Answer: This question can be solved by trigonometry ONLY.

Step-by-step explanation:

Given : PT and PS are tangent of the circle.

O is the centre of the circle and ∠SPT = 120°

To prove : OP = 2 PS.

Proof : In ΔOPS and ΔOPT

∠OSP = ∠OTP (each 90° bcz radius is perpendicular to the tangent)

PS = PT (tangent from an external point are equal)

OS = OT (radius of the circle are equal)

So,

ΔOPS ≅ ΔOPT (by SAS similarity criterion)

Therefore,

∠OPS = ∠OPT (by CPCT)

∠POS = ∠POT (by CPCT)

As,

∠SPT = 120° and ∠SOT = 60°

So, now;

∠OPS = ∠OPT = 60°

∠POS = ∠POT = 30°

Now, by taking ΔPOS

We get,

=> sin30° = PS/PO

=> 1/2 = PS/PO

=> PO = 2PS

Hence Proved.

If you did not understand this, let me know. I will explain

Answer 2

Where the two tangents PS and PT are drawn to the circle .

<SOT =120°

then OS and OT are the radii of the circle

so OS = OT

<OSP= <OTP = 90°

∆ SPO ≈ ∆PTO by SAS congruency

Thus, <POT = < POS = 120°/2= 60°

So

<POT+ <OTP+<OPT= 180°

60°+ 90°+x = 180°

<OPT = 180°-150°

<OPT = 30°

Where <OPT = 30° = <OPS

Then

In Pythagoras theorem

Op²= ps²+os²

So 4PS = 3Op